Question

Gabe deposits $2,500 into each of two savings accounts.

Account ONE earns 4% annual simple interest.
Account TWO earns 4.5% interest compounded annually.
What is the sum of the balances of Accounts ONE and TWO at the end of 3 years?

Answer 1

637.5 I think it's right

Explanation:

2,500x.04=100x3=300

2,500x.045=112.5x3=337.5

300+337.5=637.5 interest

2,500x2+637.5=5,637.5

Answer 2

$5,612.16

Explanation:

Part 1) Account I earns 4% annual simple interest.

we know that

The simple interest formula is equal to A = P (1+rt)

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest

t is Number of Time Periods

in this problem we have:

t = 3 years

P = $2,500

r = 4% = 4/100 = 0.04

substitute in the formula above

A = 2,500 (1 + 0.04 * 3)

A = 2,500 (1.12)

A = $2,800

Part 2) Account II earns 4% interest compounded annually.

we know that

The compound interest formula is equal to: A = P (1 + r/n)nt

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

t = 3 years

P = $2,500

r = 4% = 4/100 = 0.04

n = 1

substitute in the formula above

A = 2,500 * (1 + 0.04/1)3

A = 2,500 * (1.04)3

A = $2,812.16

Part 3) What is the sum of the balances of Account I and Account II at the end of 3 years?

Sum the two final investment:

$2,800 + $2,812.16 = $5,612.16